/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Solve the equation by completing... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 35

Solve the equation by completing the square. \(5 x(x+6)=-50\)

Short Answer

Expert verified
The solution to the equation is \(x = -3 + i\) and \(x = -3 - i\).

Step by step solution

01

Simplify the Equation

First, distribute the 5x into the parenthesis. This gives: \(5x^2 + 30x = -50\). Then rearrange the equation to set it to zero: \(5x^2 + 30x + 50 = 0\)
02

Factor out Constant Terms

To apply the method of completing the square, it's better to work with a coefficient of 1 for \(x^2\). To achieve this, factor out a 5 from all terms to get the equation: \(x^2 + 6x + 10 = 0\)
03

Completing the Square

To complete the square, find the number that makes \(x^2 + 6x\) a perfect square trinomial. This is done by taking half of the coefficient of x, squaring it and adding it to both sides. So, for this equation \( \frac{6}{2} = 3\) and \(3^2 = 9\). Now, add 9 to both side of the equation to obtain: \(x^2 + 6x + 9 = -1\)
04

Rewrite Equation as Perfect Square

Rewrite the left side of the equation as \((x+3)^2\). This gives the equation: \((x + 3)^2 = - 1\)
05

Solve for x

Take the square root of both sides. Don't forget that solving for x gives two solutions; one positive and one negative. This gives the solution: \(x+3 = \pm \sqrt{-1}\). This means, \(x = -3 \pm \sqrt{-1}\). The symbol indicates that there are two solutions: \(x = -3 + \sqrt{-1}\) and \(x = -3 - \sqrt{-1}\). However, the square root of -1 is undefined in the real number system, and is expressed as the imaginary number, i. Therefore, the solutions are \(x = -3 + i\) and \(x = -3 - i\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Product Property states \(\sqrt{a} \cdot \sqrt{b}=\sqrt{a b}\). Your friend concludes \(\sqrt{-4} \cdot \sqrt{-9}=\sqrt{36}=6\). Is your friend correct? Explain.

REASONING A truck that is 11 feet tall and 7 feet wide is traveling under an arch. The arch can be modeled by \(y=-0.0625 x^2+1.25 x+5.75\), where \(x\) and \(y\) are measured in feet. a. Will the truck fit under the arch? Explain. b. What is the maximum width that a truck 11 feet tall can have and still make it under the arch? c. What is the maximum height that a truck 7 feet wide can have and still make it under the arch?

Justify each step in performing the operation. $$ \begin{aligned} 11 &-(4+3 i)+5 i \\ &=[(11-4)-3 i]+5 i \\ &=(7-3 i)+5 i \\ &=7+(-3+5) i \\ &=7+2 i \end{aligned} $$

Determine whether you would use factoring, square roots, or completing the square to solve the equation. Explain your reasoning. Then solve the equation. \(3 x^2+12 x+1=0\)

Graph the function. Label the vertex, axis of symmetry, and \(x\)-intercepts. \(h(x)=2 x(x-3)\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.